You have 500 meters of fencing with which to make two enclosures. One enclosure will be in the shape of a square, and the other will have the shape of an isosceles right triangle. For example, the enclosures might look like this:

1) How long should the legs of the triangle be to minimize the combined area of the two enclosures?

The length of the legs of the triangle is approximately 86.85 meters

Step-by-step explanation:

The requirement is to minimize the combined area of the two enclosures

The given parameters are;

The length of (perimeter) fencing available = 500 meters

The shape of the first enclosure = Square

The shape of the second enclosure = Triangle

The shape of the triangle = Isosceles right triangle

The area of the square = Side, S, squared = s²

The perimeter of the square = 4 × s

The area of the triangle = 1/2 × Base × Height = 1/2 × b × b= 1/2·b²

Where;

b = The lengths of the legs of the triangle

Given that the perimeter is 500 m, we have;

length of hypotenuse of triangle = √(b² + b²) =√(2·b²) = b·√2

∴ 500 - 4·s = b + b + b·√2 = 2·b + (√2)·b = b·(2 + √2)

b = (500 - 4·s)/(2 + √2)

∴ The area of the triangle = 1/2 × ((500 - 4·s)/(2 + √2))²

The area of the square + The area of the triangle = s² + 1/2 × ((500 - 4·s)/(2 + √2))²

At minimum value, we have;

d(s² + 1/2 × ((500 - 4·s)/(2 + √2))²/ds = 0, gives (Using online application)

From which we have, s = 500/(7 + 2·√2)

Differentiating again gives;

Therefore, the point is a local minimum and the length of the legs of the triangle should be given as follows;

The length of the legs of the triangle = b = (500 - 4·s)/(2 + √2) = (500 - 4*(500/(7 + 2*√2)))/(2 + √2) ≈ 86.85

The length of the legs of the triangle ≈ 86.85 meters.